Let P(x) be a quadratic polynomial such that $$\begin{vmatrix}P(0) & P(1)\\P(0) & P(2)\end{vmatrix} = 0$$ Let P(0) = 2 and P(1) + P(2) + P(3) = 14. Then P(4) equals

Let $$P(x)=ax^2+bx+c$$, and it is given that P(0) = 2.

=> $$a(0)+b(0)+c=2$$ => $$c=2$$

$$\begin{vmatrix}P(0) & P(1)\\P(0) & P(2)\end{vmatrix} = 0$$

=> $$P\left(0\right)P\left(2\right)-P\left(0\right)P\left(1\right)=0$$

=> $$P(1)=P(2)$$

=> $$a+b+2=4a+2b+2$$

=> $$b=-3a$$

It is also given that -

$$P(1) + P(2) + P(3) = 14$$

$$a+b+2+4a+2b+2+9a+3b+2=14$$

$$14a+6b=8$$                (Substituting the value of b = -3a)

$$14a-18a=8$$ => $$a=-2$$

Thus, $$b=-3(-2)=6$$

Therefore, $$P(x)=-2x^2+6x+2$$

=> $$P(4)=-2*4^2+6*4+2$$

=> $$P(4)=-32+24+2$$

=> $$P(4)=-6$$